asst 7

Am I as far behind as I think? Thus far, I have done all the textbook problems through Section R7. I needed the practice. If I am excessively overdue, I may have to begin doing just the problems assigned.

You are a bit behind. However the quality of your work is very good. I'd rather have you work slowly and well than work fast and fail to pick up things you'll need later. If you don't get the course done by the end of the semester, but have passed tests on Chapters R, 1 and 2, then I would give you an Incomplete and extra time to finish. But try to be sure you're on track to do at least that.

assignment #007 ƏNyZ߻ޟDJ College Algebra 10-14-2005

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09:15:25 Query R.7.10 (was R.7.6). Show how you reduced (x^2 + 4 x + 4) / (x^4 - 16) to lowest terms.

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RESPONSE --> (x^2 + 4 x + 4) / (x^4 - 16) = (x + 2) (x + 2) / (x^2 - 4) (x^2 + 4) = (x + 2) (x + 2) / (x - 2) (x + 2) (x^2 + 4) = (x + 2) / (x - 2) (x^2 + 4)

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09:49:08 ** We factor the denominator to get first (x^2-4)(x^2+4), then (x-2)(x+2)(x^2+4). The numerator factors as (x+2)^2. So the fraction is (x+2)(x+2)/[(x-2)(x+2)(x^2+4)], which reduces to (x+2)/[(x-2)(x^2+4)]. **

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RESPONSE --> ok

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09:59:26 Query R.7.28 (was R.7.24). Show how you simplified[ ( x - 2) / (4x) ] / [ (x^2 - 4 x + 4) / (12 x) ].

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RESPONSE --> [ ( x - 2) / (4x) ] / [ (x^2 - 4 x + 4) / (12 x) ] = [ ( x - 2) / (4x) ] / [ (x -2 ) (x - 2) / (12 x) ] = [ ( x - 2) / (4x) ] * [ (12 x) / (x -2 ) (x - 2) ] = [ ( x - 2 ) ( 12x) ] / [ (4x) (x -2 ) (x - 2) ] = (12x) / [ (4x) ( x - 2)] = (3 * 4x) / [ (4x) ( x - 2)] = 3 / x - 2

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09:59:35 ** [ ( x - 2) / (4x) ] / [ (x^2 - 4 x + 4) / (12 x) ] = (x-2) * / 4x * 12 x / (x^2 - 4x + 4) = (x-2) * 12 x / [ 4x ( x^2 - 4x + 4) ] = 12 x (x-2) / [4x ( x-2) ( x-2) ] = 3/(x - 2) **

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RESPONSE --> ok

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10:06:32 Query R.7.40 (was R.7.36). Show how you found and simplified the sum (2x - 5) / (3x + 2) + ( x + 4) / (3x + 2).

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RESPONSE --> (2x - 5) / (3x + 2) + ( x + 4) / (3x + 2) = (2x - 5) + ( x + 4) / (3x + 2) = 3x - 1 / 3x + 2

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10:06:41 ** We have two like terms so we write (2x-5)/(3x+2) + (x+4)/(3x+2) = [(2x-5)+(x+4)]/(3x+2). Simplifying the numerator we have (3x-1)/(3x+2). **

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RESPONSE --> ok

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10:30:04 Query R.7.52 (was R.7.48). Show how you found and simplified the expression(x - 1) / x^3 + x / (x^2 + 1).

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RESPONSE --> LCM x^3 (x + 1) (x - 1) (x - 1) / x^3 + x / (x^2 + 1) = [ (x - 1) (x + 1) (x - 1) ] / [ (x^3) (x + 1) (x - 1) ] + [ (x) (x^3) ] / [ ( x^3) (x + 1) (x - 1) ] = [ (x - 1) (x + 1) (x - 1) + (x) (x^3) ] / [ ( x^3) (x + 1) (x - 1) ] = [ (x - 1)^2 (x + 1) + (x^4) ] / [x^3 ( x + 1) (x - 1) ] = ( x^4 + x - 1 ) / x^3

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10:34:40 ** Starting with (x-1)/x^3 + x/(x^2+1) we multiply the first term by (x^2 + 1) / (x^2 + 1) and the second by x^3 / x^3 to get a common denominator: [(x-1)/(x^3) * (x^2+1)/(x^2+1)]+[(x)/(x^2+1) * (x^3)/(x^3)], which simplifies to (x-1)(x^2+1)/[ (x^3)(x^2+1)] + x^4/ [(x^3)(x^2+1)]. Since the denominator is common to both we combine numerators: (x^3+x-x^2-1+x^4) / ) / [ (x^3)(x^2+1)] . We finally simplify to get (x^4 +x^3 - x^2+x-1) / ) / [ (x^3)(x^2+1)] **

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RESPONSE --> ok

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10:42:22 Query R.7.58 (was R.7.54). How did you find the LCM of x - 3, x^3 + 3x and x^3 - 9x, and what is your result?

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RESPONSE --> x - 3 = (x-3) x^3 + 3x = x (x^2+3) x^3 - 9x = x (x-3) (x+3) LCM = x (x-3) (x+3) (x^2+3)

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10:45:04 ** x-3, x^3+3x and x^3-9x factor into x-3, x(x^2+3) and x(x^2-9) then into (x-3) , x(x^2+3) , x(x-3)(x+3). The factors x-3, x, x^2 + 3 and x + 3 'cover' all the factors of the three polynomials, and all are needed to do so. The LCM is therefore: x(x-3)(x+3)(x^2+3) **

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RESPONSE --> ok

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10:50:37 Query R.7.64 (was R.7.60). Show how you found and simplified the difference3x / (x-1) - (x - 4) / (x^2 - 2x + 1).

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RESPONSE --> 3x / (x-1) - (x - 4) / (x^2 - 2x + 1) LCM = 3x (x-1) / (x-1) (x-1) - (x-4) / (x-1) (x-1) = 3x^2 - 3x - x - 4 / (x-1) (x-1) = 3x^2 - 4x - 4 / (x-1)^2

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10:50:53 ** Starting with 3x / (x-1) - (x-4) / (x^2 - 2x +1) we factor the denominator of the second term to obtain (x - 1)^2. To get a common denominator we multiply the first expression by (x-1) / (x-1) to get 3x(x-1)/(x-1)^2 - (x-4)/(x-1)^2, which gives us (3x^2-3x-x-4) / (x-1)^2 = (3x^2 - 4x - 4) / (x-1)^2. DRV**

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RESPONSE --> ok

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10:52:10 QUESTION FROM STUDENT: On the practice test I'm having problems with problem #5 I don't know where to start or how to set it up. I'm probably missing something simple and will probably feel stupid by seeing the solution. Could you help with this problem. A retailer is offering 35% off the purchase price of any pair of shoes during its annual charity sale. The sale price of the shoes pictured in the advertisement is $44.85. Find the original price of the shoes by solving the equation p-.35p = 44.85 for p. INSTRUCTOR RESPONSE: It's very easy to get ahold of the wrong idea on a problem and then have trouble shaking it, or to just fail to look at it the right way. Nothing stupid about it, just human nature. See if the following makes sense. If not let me know. p - .35 p = 44.85. Since p - .35 p = 1 p - .35 p = (1 - .35) p = .65 p we have .65 p = 44.85. Multiplying both sides by 1/.65 we get p = 44.85 / .65 = etc. (you can do the division on your calculator); you'll get something near $67).

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RESPONSE --> OK

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